Finished all chapters and definitions. I need to add subsections and see if there's any theorems or definitions in the appendicies that are worth adding to this as well.

chapter-1/bases-and-dimension.tex

@@ -8,9 +8,9 @@
 \begin{theorem}
 	\hfill\\
 	Let $V$ be a vector space and $\beta = \{v_1, v_2, \dots, v_n\}$ be a subset of $V$. Then $\beta$ is a basis for $V$ if and only if each $v \in V$ can be uniquely expressed as a linear combination of vectors of $\beta$, that is, can be expressed in the form
-	
+
 	\[v = a_1v_1 + a_2v_2 + \dots + a_nv_n\]
-	
+
 	for unique scalars $a_1, a_2, \dots, a_n$.
 \end{theorem}
 
@@ -38,9 +38,9 @@
 	Let $V$ be a vector space with dimension $n$.
 	\begin{enumerate}
 		\item Any finite generating set for $V$ contains at least $n$ vectors, and a generating set for $V$ that contains exactly $n$ vectors is a basis for $V$.
-		
+
 		\item Any linearly independent subset of $V$ that contains exactly $n$ vectors is a basis for $V$.
-		
+
 		\item Every linearly independent subset of $V$ can be extended to a basis for $V$.
 	\end{enumerate}
 \end{corollary}
@@ -59,49 +59,49 @@
 \begin{definition}[\textbf{The Lagrange Interpolation Formula}]
 	\hfill\\
 	Corollary 2 of the replacement theorem can be applied to obtain a useful formula. Let $c_0, c_1, \dots, c_n$ be distinct scalars in an infinite field $\F$. The polynomials $f_0(x), f_1(x), \dots, f_n(x)$ defined by
-	
+
 	\[f_i(x) = \frac{(x-c_0)\dots(x-c_{i-1})(x-c_{i+1})\dots(x-c_n)}{(c_i - c_0)\dots(c_i-c_{i-1})(c_i-c_{i+1})\dots(c_i-c_n)} = \prod_{\substack{k=0 \\ k \neq i}}^{n} \frac{x-c_k}{c_i - c_k}\]
-	
-	are called the \textbf{Lagrange polynomials} (associated with $c_0, c_1, \dots, c_n$). Note that each $f_i(x)$ is a polynomial of degree $n$ and hence is in $P_n(\F)$. By regarding $f_i(x)$ as a polynomial function $f_i: \F \to \F$, we see that 
-	
+
+	are called the \textbf{Lagrange polynomials} (associated with $c_0, c_1, \dots, c_n$). Note that each $f_i(x)$ is a polynomial of degree $n$ and hence is in $P_n(\F)$. By regarding $f_i(x)$ as a polynomial function $f_i: \F \to \F$, we see that
+
 	\begin{equation}
 		f_i(c_j) = \begin{cases}
-			0 &\text{if}\ i \neq j,\\
-			1 &\text{if}\ i = j.
+			0 & \text{if}\ i \neq j, \\
+			1 & \text{if}\ i = j.
 		\end{cases}
 	\end{equation}
-	
-	This property of Lagrange polynomials can be used to show that $\beta = \{f_0, f_1, \dots, f_n\}$ is a linearly independent subset of $P_n(\F)$. Suppose that 
-	
+
+	This property of Lagrange polynomials can be used to show that $\beta = \{f_0, f_1, \dots, f_n\}$ is a linearly independent subset of $P_n(\F)$. Suppose that
+
 	\[\sum_{i=0}^{n}a_if_i = 0\ \ \text{for some scalars}\ a_0, a_1, \dots, a_n,\]
-	
+
 	where $0$ denotes the zero function. Then
-	
+
 	\[\sum_{i=0}^{n}a_if_i(c_j)=0\ \ \text{for}\ j=0, 1, \dots, n.\]
-	
+
 	But also
-	
+
 	\[\sum_{i=0}^{n}a_if_i(c_j)=a_j\]
-	
+
 	by (1.1). Hence $a_j = 0$ for $j = 0, 1, \dots, n$; so $\beta$ is linearly independent. Since the dimension of $P_n(\F)$ is $n + 1$, it follows from Corollary 2 of the replacement theorem that $\beta$ is a basis for $P_n(\F)$.
-	
+
 	Because $\beta$ is a basis for $P_n(\F)$, every polynomial function $g$ in $P_n(\F)$ is a linear combination of polynomial functions of $\beta$, say,
-	
+
 	\[g = \sum_{i=0}^{n}b_if_i.\]
-	
-	It follows that 
-	
+
+	It follows that
+
 	\[g(c_j)=\sum_{i=0}^{n}b_if_i(c_j)=b_j;\]
-	
+
 	so
-	
+
 	\[g=\sum_{i=0}^{n}g(c_i)f_i\]
-	
+
 	is the unique representation of $g$ as a linear combination of elements of $\beta$. This representation is called the \textbf{Lagrange interpolation formula}. Notice that the preceding argument shows that if $b_0, b_1, \dots, b_n$ are any $n+1$ scalars in $\F$ (not necessarily distinct), then the polynomial function
-	
+
 	\[g = \sum_{i=0}^{n}b_if_i\]
-	
+
 	is the unique polynomial in $P_n(\F)$ such that $g(c_j) = b_j$. Thus we have found the unique polynomial of degree not exceeding $n$ that has specified values $b_j$ at given points $c_j$ in its domain ($j = 0, 1, \dots, n$).\\
-	
+
 	An important consequence of the Lagrange interpolation formula is the following result: If $f \in P_n(\F)$ and $f(c_i) = 0$, for $n+1$ distinct scalars $c_0, c_1, \dots, c_n$ in $\F$, then $f$ is the zero function.
-\end{definition}
+\end{definition}